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The opposite category of the category of rings is equivalent to the category of affine schemes, via the Spec functor.

Is there a similar result if we consider the Proj construction, that takes a graded ring and returns some scheme?

This is a question from a beginner in algebraic geometry, trying to understand how things fit together.

Bruno Stonek
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    Thrre are several things you can do to your graded ring without changing its Proj. Taking veronese subrings, for example: you keep only the graded components of degree divisible by 17, say —this is a graded subring with the same Proj. – Mariano Suárez-Álvarez Nov 01 '13 at 17:35
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    @Mariano: Hi :) is there at least something functorially interesting to be said about this construction? – Bruno Stonek Nov 01 '13 at 17:45
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    @BrunoStonek It isn't functorial. A map of graded rings $\phi:R\to S$ induces a map $\text{Proj}(S)/V(\phi(R_+))\to \text{Proj}(S)$, not necessarily a map on full proj's. In fact, in a way that can be made precise, it's non-functorial in the same way that taking orbits in the category of $G$-sets isn't functorial. – Alex Youcis Nov 01 '13 at 19:15
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    Dear Alex, @AlexYoucis I think this can be overcome by considering a smaller category: graded rings with homomorphisms $\phi : R\to S$ satisfying the property $\sqrt{\phi(R_+)S}\supseteq S_+$. Now given a projective scheme $X$, one can ask whether inside the $R$ such that $\mathrm{Proj}(R)\cong X$, there exists a canonical one ? In other word, one can ask whether there exists a final or initial graded ring among these $R$ (homomorphisms are those with the above restriction). The answer is unfortunately no: $R$ has the same Proj as $\oplus_n R_{dn}$ for any $d\ge 2$ (see Mariano's comment). ... – Cantlog Nov 02 '13 at 09:48
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    On the other hand, for example if $R=k[T_0,\dots, T_n]/I$, then $k[S, T_0, \dots, T_n]/(S^2, ST_0, \dots, ST_n)$ has the same Proj, but is bigger than $R$. – Cantlog Nov 02 '13 at 09:49
  • @Cantlog So, you're basically saying take a subcategory of graded rings where you basically force $V(\phi(R_+))=\varnothing$? That certainly fixes non-functorality, but besides this fix, is there a natural reason to consider these maps? – Alex Youcis Nov 02 '13 at 09:53
  • @AlexYoucis: Yes. No, I don't know, but this seems to be indispensable if we want to study functoriality. – Cantlog Nov 02 '13 at 09:57
  • @Alex: To have a category of some sort of structure, you need to consider structure-preserving maps. In some sense, the collection of homomorphisms is what defines the actual structure on your objects, and if you pick the wrong collection of homomorphisms, then the structure isn't going to be what you thought you were studying. –  Nov 02 '13 at 15:13

3 Answers3

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This question is over a year old, but I have a different perspective which might interest you.

As others have pointed out, $\text{Proj}$ is not even a functor, let alone fully faithful, but it does factor through an equivalence in the following sense. A $\mathbb{Z}$-grading on a ring $A$ is the same as an action of the multiplicative group $\mathbb{G}_m$ on $\text{Spec } A$. Homomorphisms of graded rings correspond to $\mathbb{G}_m$-equivariant morphisms of schemes. So $\text{Spec}$ induces an equivalence from graded rings to affine schemes with $\mathbb{G}_m$-action.

But $\text{Proj}$ is only defined on nonnegatively graded rings. Call a $\mathbb{G}_m$-action on scheme $X$ contracting if it extends to an action of the multiplicative monoid $\mathbb{A}^1$. Intuitively, this means that the $\mathbb{G}_m$-action contracts $X$ onto its fixed point locus $X^{\mathbb{G}_m}$. Then a graded ring $A$ is nonnegatively graded if and only if the corresponding $\mathbb{G}_m$-action on $\text{Spec } A$ is contracting.

So we can interpret $\text{Proj}$ as a construction which builds a scheme from an affine scheme with contracting $\mathbb{G}_m$-action. Namely, it sends $X$ to the GIT quotient $(X \setminus X^{\mathbb{G}_m})//\mathbb{G}_m$.

The failure of functoriality is due to the fact that a morphism $X \to Y$ can send points in $X \setminus X^{\mathbb{G}_m}$ to points in $Y^{\mathbb{G}_m}$. There is also some loss of information because any nonzero power of the $\mathbb{G}_m$-action produces the same GIT quotient: this is the Veronese stuff.

Justin Campbell
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Unfortunately, there is not. As Mariano points out, non-isomorphic rings might happen to have isomorphic Proj. However, the biggest problem is pointed out by Alex: Proj is not even a functor, unless you add some restrictions. The problem is already illustrated at the level of vector spaces. If $V \to W$ is a map of vector spaces, you do not get a map $P(V) \to P(W)$ from it, unless it is an injection. Indeed, where would you send lines in the kernel?

Bruno Joyal
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I'm not sure about proj but if you think about not just rings but modules over those rings then you get some results. First let us look at affine schemes. We have the usual $$\operatorname{Hom}(A,B) = \operatorname{Hom}(\operatorname{Spec} B,\operatorname{Spec}A),$$

i.e. there is an equivalence of categories between affine schemes and commutative unital rings. Now you can ask the following. If I study modules over a ring, is there a "module-theoretic" category to which this is equivalent to? Yes! Your answer is given by the category of quasi-coherent sheaves on a space $X$ denoted $\operatorname{QCoh}(X)$. Now if $X = \operatorname{Spec} R$ is affine, there is an equivalence of categories

$$\textbf{$R$-Mod} \leftrightarrow \textbf{QCoh}(X)$$ where the equivalence is given by $M \mapsto \widetilde{M}$ and for the other way we send a quasi-coherent sheaf to its global sections. One can also prove the analogous (though arrow preserving!) result $$\operatorname{Hom}_R(M,N) = \operatorname{Hom}_{\mathcal{O}_X}(\widetilde{M},\widetilde{N})$$

where the R.H.S. is global Hom and not sheaf hom.

Now onto your question about Proj. I don't know of a result involving purely proj but if you look at "modules over rings" we have:

Exercise 2.5.9 Hartshorne: Let $S$ be a graded ring, generated in degree $1$ as an $S_0$-algebra. Let $M$ be a graded $S$ module and $X = \operatorname{Proj} S$. There is an equivalence of cateogires between the category of f.g. graded $S$-modules (modulo the equivalence that $M\sim M'$ if they agree in sufficiently high degree) and coherent sheaves on $X$.

  • Thank you for your answer. I have not yet studied modules or quasi-coherent things, but when I do I (it should be shortly) I will come back to this. – Bruno Stonek Nov 03 '13 at 16:50